TITLE: can a Linear operator on infinite dimensional vector space have infinite eigenvalues?
QUESTION [5 upvotes]: I know that for a finite dimensional vector space, the number of eigenvalues is at most the dimension of the vector space. Is there an example of an infinite dimensional vector space and a linear operator such that the linear operator allows infinite eigenvalues? I cannot think of such an example, but I was wondering if it is possible to construct one since I was also unable to prove why the number of eigenvalues would necessarily be finite in such a case.

REPLY [4 votes]: Another example from a different area just for the sake of building intuition. If you let $C^1(\mathbb{R})$ be the vector space of continuously differentiable functions $f:\mathbb{R} \to \mathbb{R}$, and let $L$ be the derivative operator, mapping $L(f) = f'$, then the eigenvalues of $L$ satisfy
$$
kf = L(f) = f',
$$
in other words any real $k$ is an eigenvalue with the eigenvector
$$
f(x) = e^{kx}.
$$
So here is an example with not just infinite, but uncountable eigenvectors/eigenvalues.